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Homework Help: Level of water rising in a cone.

  1. Jun 30, 2012 #1
    1. The problem statement, all variables and given/known data
    A right circular cone with radius r and height h is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the cone is half full.

    2. Relevant equations

    3. The attempt at a solution
    V=5t. The level of the water is determined by h, so I need the find the derivative of h. h=15t/∏r2, but I'm pretty sure that's not right because r is not a constant, it depends on h. That's where I'm stuck.
  2. jcsd
  3. Jun 30, 2012 #2


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    Looks like some info is missing in the problem. For instance, if we knew the relationship between r and h, then we can rewrite r in terms of h and get rid of the r. Can you double check and make sure you have the entire problem?
  4. Jun 30, 2012 #3
    That's what I was thinking too. Here's the question in it's entirety:
    "A cup in the form of a right circular cone with radius r and height h is being filled with water at the rate of 5 cu in./sec. How fast is the level of the water rising when the volume of the water is equal to one half the volume of the cup?"
  5. Jun 30, 2012 #4
    Find r in term of h.
  6. Jul 1, 2012 #5
    How can I do that if I don't know the angle between h and the slant height.
  7. Jul 1, 2012 #6
    Here's my (attempted) solution given the information I have.
    r=htanθ, θ is the angle between h and the slant height.
    V=5t=tan2θh3∏/3, therefore h=(15t/∏tan2θ)1/3
    and h'=(5/∏tan2θ)*(∏tan2θ/15t)2/3.
    When the cup is half full t=tan2h3∏/30 and
    Last edited: Jul 1, 2012
  8. Jul 1, 2012 #7


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    It is correct if you add the parentheses, but you can replace tanθ=r/h and get the result in the given radius of the cone.

    Last edited: Jul 1, 2012
  9. Jul 1, 2012 #8
    Consider the cone (or at least the cross-section of it) on the x-y plane, such that the point is at the origin and the open end is at x=h, y=-r to +r.

    For any x (from 0 to h), you know:

    y = ?

    V = ?

    dV/dt = ? (this will involve a dx/dt term) - and you know this equals 5

    So you want dx/dt, where x solves V(x)=(∏r2h/3)/2

    I'm not sure if this is the simplest way, but that's how I'd approach it.

    (Hope I'm right!)
  10. Jul 1, 2012 #9


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    I would normally prefer to let the constant dimensions of the cone be represented by the big letters, but I'm stuck because the problem used small letters. Never mind.

    Let R = R(t) and H = H(t) represent the instantaneous radius of the circular surface of the water and the instantaneous height of the water above the vertex, respectively.

    Use similar triangles to determine the relationship between R and H in terms of the constant radius r and height h of the conical container. You don't need trig.

    Once that's done, use [itex]V = \frac{1}{3}{\pi}R^2H[/itex] to find an expression for V = V(t) in terms of R and H. That represents the instantaneous volume of water at time t. Use the previously derived relationship between R and H to express V only in terms of H. In other words, you now have V as a function of H, i.e. V = f(H).

    Now find [itex]\frac{dV}{dH}[/itex]. By Chain Rule, you know that [itex]\frac{dV}{dt} = \frac{dV}{dH}.\frac{dH}{dt}[/itex], so [itex]\frac{dH}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dH}}[/itex]. Put the expression for [itex]\frac{dV}{dH}[/itex] into the denominator, so that there will be a H term in the RHS of the equation.

    You're already given [itex]\frac{dV}{dt}[/itex] (constant at 5 cu in/sec.), so all you need now is to find the value of H (using V = f(H)) when the cone is half full (i.e. the volume of water is half the volume of the container), put it all into the last equation and compute.
  11. Jul 1, 2012 #10


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    It needs some parentheses: h'=5*41/3/(∏tan2θh2)

    Have you all got different result from that of the OP? I think it is correct.

  12. Jul 1, 2012 #11
    @Curious3141 How would you relate r to h without using trig, cause I see no other way, though I could be missing something.

    @ehild You're right I forgot some parentheses. Fixed it.
  13. Jul 1, 2012 #12


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    I'm using big R and big H for the "water cone" and small r and small h for the conical container. Of course, the dimensions of the latter are bigger.

    Sketch a vertical cross-section through the middle of the cone. The container will be an isosceles triangle made up of two identical adjacent right triangles. The shape of the water will just be a scaled down version of the same, made up of two smaller identical right triangles. Can you see this?

    Then, by similar triangles, R/H = r/h. That's it.
  14. Jul 1, 2012 #13
    I see what you mean, R then equals Hh/r which eliminates the need for the angle θ. I kept getting confused because there are two values for both the radius and the height, one is a constant and the other is a variable.
  15. Jul 1, 2012 #14


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    That should be [itex]R = \frac{rH}{h}[/itex].

    Which is why I prefer to use the small letters for the variables and the big letters for the constants. If we did this, then [itex]r \leq R[/itex] and [itex]h \leq H[/itex], where the small letters refer to the variables describing the water, and the big ones the constants describing the cone. I think you'll find this far more natural.

    Unfortunately, the question used small letters for the cone, so we have to make do with the opposite arrangement.
  16. Jul 1, 2012 #15


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    Similar triangles. If the cone has radius R and height H, then the radius, r, of the circular top of water that fills the cone to height h satisfies r/h= R/H because they are "corresponding parts of similar triangles".

  17. Jul 1, 2012 #16
    @Curious3141 Once again you're right. I really need to get some sleep.

    Thanks for the help guys.
  18. Jul 2, 2012 #17
    Hi ehild,

    Yes, I think we're all singing from the same hymn sheet. :smile:
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